Divisibile by 7 theorem check
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I have come up with a theorem to check if a number is divisible by 7.
A number is $overlineabcdefg$ will be divisible by 7 if the sum below is divisible by 7
$$(-2)^0a + (-2)^1b + (-2)^2c + (-2)^3d + (-2)^4e + (-2)^5f + (-2)^6g$$
It is valid for any number of digits..
I can prove it mathematically..
So I need your help to check if this theorem aren’t present/known to the world or not.
If not, what can be done to publish it.
number-theory
edited Aug 21 at 13:14
Stefan4024
29.2k53276
asked Aug 21 at 13:07
Ravi Goyal
161
 |Â
show 5 more comments
up vote
3
down vote
favorite
I have come up with a theorem to check if a number is divisible by 7.
A number is $overlineabcdefg$ will be divisible by 7 if the sum below is divisible by 7
$$(-2)^0a + (-2)^1b + (-2)^2c + (-2)^3d + (-2)^4e + (-2)^5f + (-2)^6g$$
It is valid for any number of digits..
I can prove it mathematically..
So I need your help to check if this theorem aren’t present/known to the world or not.
If not, what can be done to publish it.
number-theory
edited Aug 21 at 13:14
Stefan4024
29.2k53276
asked Aug 21 at 13:07
Ravi Goyal
161
1
What do $a,b,c,ldots$ represent? You haven't defined them. Without understanding your claim competely, I would suggest that you do not have your hopes up. Such rules are very well-understood for hundreds of years, these are not publishable.
– A. Pongrácz
Aug 21 at 13:11
If you believe the result is original, it could have been a good ideia to draft a paper and submit it to arxiv.org where some mathematicians publish time-sensitive results to ensure they're able to claim authorship. Then submit it to a good publication, where experts should check the validity and originality of your claim. Be careful that not much attention needs to be given if a poorly written paper is submitted. Good literature review is also important.
– Mefitico
Aug 21 at 13:12
Does this check only work for $7$-digit numbers?
– 5xum
Aug 21 at 13:15
It is as cumbersome as dividing a digit by $7$. Divisibility tests tend to make it easier like in the case of $2,3,11$ etc. Do you think this would be practical to solve for a $20$ digit number?
– prog_SAHIL
Aug 21 at 13:17
2
This is essentially the divisibility test on the wiki page where you double the last digit and subtract it from the number formed by the other digits. Then repeat the process with this new number, and so on until you have a single digit left. This is almost the same as multiplying the digits by powers of $-2$ as you are doing, but easier.
– Jaap Scherphuis
Aug 21 at 13:34
 |Â
show 5 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have come up with a theorem to check if a number is divisible by 7.
A number is $overlineabcdefg$ will be divisible by 7 if the sum below is divisible by 7
$$(-2)^0a + (-2)^1b + (-2)^2c + (-2)^3d + (-2)^4e + (-2)^5f + (-2)^6g$$
It is valid for any number of digits..
I can prove it mathematically..
So I need your help to check if this theorem aren’t present/known to the world or not.
If not, what can be done to publish it.
number-theory
edited Aug 21 at 13:14
Stefan4024
29.2k53276
asked Aug 21 at 13:07
Ravi Goyal
161
I have come up with a theorem to check if a number is divisible by 7.
A number is $overlineabcdefg$ will be divisible by 7 if the sum below is divisible by 7
$$(-2)^0a + (-2)^1b + (-2)^2c + (-2)^3d + (-2)^4e + (-2)^5f + (-2)^6g$$
It is valid for any number of digits..
I can prove it mathematically..
So I need your help to check if this theorem aren’t present/known to the world or not.
If not, what can be done to publish it.
number-theory
edited Aug 21 at 13:14
Stefan4024
29.2k53276
asked Aug 21 at 13:07
Ravi Goyal
161
edited Aug 21 at 13:14
Stefan4024
29.2k53276
edited Aug 21 at 13:14
Stefan4024
29.2k53276
edited Aug 21 at 13:14
Stefan4024
29.2k53276
29.2k53276
asked Aug 21 at 13:07
Ravi Goyal
161
asked Aug 21 at 13:07
Ravi Goyal
161
asked Aug 21 at 13:07
Ravi Goyal
161
161
1
What do $a,b,c,ldots$ represent? You haven't defined them. Without understanding your claim competely, I would suggest that you do not have your hopes up. Such rules are very well-understood for hundreds of years, these are not publishable.
– A. Pongrácz
Aug 21 at 13:11
If you believe the result is original, it could have been a good ideia to draft a paper and submit it to arxiv.org where some mathematicians publish time-sensitive results to ensure they're able to claim authorship. Then submit it to a good publication, where experts should check the validity and originality of your claim. Be careful that not much attention needs to be given if a poorly written paper is submitted. Good literature review is also important.
– Mefitico
Aug 21 at 13:12
Does this check only work for $7$-digit numbers?
– 5xum
Aug 21 at 13:15
It is as cumbersome as dividing a digit by $7$. Divisibility tests tend to make it easier like in the case of $2,3,11$ etc. Do you think this would be practical to solve for a $20$ digit number?
– prog_SAHIL
Aug 21 at 13:17
2
This is essentially the divisibility test on the wiki page where you double the last digit and subtract it from the number formed by the other digits. Then repeat the process with this new number, and so on until you have a single digit left. This is almost the same as multiplying the digits by powers of $-2$ as you are doing, but easier.
– Jaap Scherphuis
Aug 21 at 13:34
 |Â
show 5 more comments
1
What do $a,b,c,ldots$ represent? You haven't defined them. Without understanding your claim competely, I would suggest that you do not have your hopes up. Such rules are very well-understood for hundreds of years, these are not publishable.
– A. Pongrácz
Aug 21 at 13:11
If you believe the result is original, it could have been a good ideia to draft a paper and submit it to arxiv.org where some mathematicians publish time-sensitive results to ensure they're able to claim authorship. Then submit it to a good publication, where experts should check the validity and originality of your claim. Be careful that not much attention needs to be given if a poorly written paper is submitted. Good literature review is also important.
– Mefitico
Aug 21 at 13:12
Does this check only work for $7$-digit numbers?
– 5xum
Aug 21 at 13:15
It is as cumbersome as dividing a digit by $7$. Divisibility tests tend to make it easier like in the case of $2,3,11$ etc. Do you think this would be practical to solve for a $20$ digit number?
– prog_SAHIL
Aug 21 at 13:17
2
This is essentially the divisibility test on the wiki page where you double the last digit and subtract it from the number formed by the other digits. Then repeat the process with this new number, and so on until you have a single digit left. This is almost the same as multiplying the digits by powers of $-2$ as you are doing, but easier.
– Jaap Scherphuis
Aug 21 at 13:34
1
1
What do $a,b,c,ldots$ represent? You haven't defined them. Without understanding your claim competely, I would suggest that you do not have your hopes up. Such rules are very well-understood for hundreds of years, these are not publishable.
– A. Pongrácz
Aug 21 at 13:11
What do $a,b,c,ldots$ represent? You haven't defined them. Without understanding your claim competely, I would suggest that you do not have your hopes up. Such rules are very well-understood for hundreds of years, these are not publishable.
– A. Pongrácz
Aug 21 at 13:11
If you believe the result is original, it could have been a good ideia to draft a paper and submit it to arxiv.org where some mathematicians publish time-sensitive results to ensure they're able to claim authorship. Then submit it to a good publication, where experts should check the validity and originality of your claim. Be careful that not much attention needs to be given if a poorly written paper is submitted. Good literature review is also important.
– Mefitico
Aug 21 at 13:12
If you believe the result is original, it could have been a good ideia to draft a paper and submit it to arxiv.org where some mathematicians publish time-sensitive results to ensure they're able to claim authorship. Then submit it to a good publication, where experts should check the validity and originality of your claim. Be careful that not much attention needs to be given if a poorly written paper is submitted. Good literature review is also important.
– Mefitico
Aug 21 at 13:12
Does this check only work for $7$-digit numbers?
– 5xum
Aug 21 at 13:15
Does this check only work for $7$-digit numbers?
– 5xum
Aug 21 at 13:15
It is as cumbersome as dividing a digit by $7$. Divisibility tests tend to make it easier like in the case of $2,3,11$ etc. Do you think this would be practical to solve for a $20$ digit number?
– prog_SAHIL
Aug 21 at 13:17
It is as cumbersome as dividing a digit by $7$. Divisibility tests tend to make it easier like in the case of $2,3,11$ etc. Do you think this would be practical to solve for a $20$ digit number?
– prog_SAHIL
Aug 21 at 13:17
2
2
This is essentially the divisibility test on the wiki page where you double the last digit and subtract it from the number formed by the other digits. Then repeat the process with this new number, and so on until you have a single digit left. This is almost the same as multiplying the digits by powers of $-2$ as you are doing, but easier.
– Jaap Scherphuis
Aug 21 at 13:34
This is essentially the divisibility test on the wiki page where you double the last digit and subtract it from the number formed by the other digits. Then repeat the process with this new number, and so on until you have a single digit left. This is almost the same as multiplying the digits by powers of $-2$ as you are doing, but easier.
– Jaap Scherphuis
Aug 21 at 13:34
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
This is a trivial assertion if you understand modular arithmetic, which is absolutely basic in number theory.
Note that $1/(-2)equiv 10 pmod 7$, which implies your claim immediately.
However, this rule has disadvantages compared to the usual algorithm that decides if a number is divisible by $7$ or not (namely, break it up to $3$-digit numbers, add them with alternating signs, iterate...)
- Your method only decides if the number is divisible by $7$, it does not return the remainder $pmod 7$
- Your method is very slow, it takes much more time to compute that expression than adding up $3$-digit numbers.
So I suggest you read about advanced number theory, and if you find an interesting open problem (that experts found interesting as well), work on it.
But this result is unpublishable.
In any case, congratulations for finding this method.
answered Aug 21 at 13:20
A. Pongrácz
4,112625
I was hoping that the OP would take my hint & discover the 3 digit method for himself. But I guess it's still a good exercise to prove why it works.
– PM 2Ring
Aug 21 at 13:31
I think I have already finished my post (or maybe still typing it) when you made your suggestion, so I didn't see it. Sorry for that.
– A. Pongrácz
Aug 21 at 13:38
Not a problem! Your answer is fine.
– PM 2Ring
Aug 21 at 13:44
I am not sure if I would say it is trivial. The reversal of the order of the digits when passing from base 10 to base -2 is a nice twist which looks somewhat surprising. But it is true: things like these have been studied before in more generality. For instance you (that is: @Ravi Goyal, not A. Pongracz) could look at this page en.wikipedia.org/wiki/Reciprocal_polynomial where the assertion from the current answer that '[the fact that $1/(-2) = 10 mod 7$] implies your claim immediately' is (in slightly more general notation) listed as 'Property 2.'
– Vincent
Aug 21 at 15:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
This is a trivial assertion if you understand modular arithmetic, which is absolutely basic in number theory.
Note that $1/(-2)equiv 10 pmod 7$, which implies your claim immediately.
However, this rule has disadvantages compared to the usual algorithm that decides if a number is divisible by $7$ or not (namely, break it up to $3$-digit numbers, add them with alternating signs, iterate...)
- Your method only decides if the number is divisible by $7$, it does not return the remainder $pmod 7$
- Your method is very slow, it takes much more time to compute that expression than adding up $3$-digit numbers.
So I suggest you read about advanced number theory, and if you find an interesting open problem (that experts found interesting as well), work on it.
But this result is unpublishable.
In any case, congratulations for finding this method.
answered Aug 21 at 13:20
A. Pongrácz
4,112625
I was hoping that the OP would take my hint & discover the 3 digit method for himself. But I guess it's still a good exercise to prove why it works.
– PM 2Ring
Aug 21 at 13:31
I think I have already finished my post (or maybe still typing it) when you made your suggestion, so I didn't see it. Sorry for that.
– A. Pongrácz
Aug 21 at 13:38
Not a problem! Your answer is fine.
– PM 2Ring
Aug 21 at 13:44
I am not sure if I would say it is trivial. The reversal of the order of the digits when passing from base 10 to base -2 is a nice twist which looks somewhat surprising. But it is true: things like these have been studied before in more generality. For instance you (that is: @Ravi Goyal, not A. Pongracz) could look at this page en.wikipedia.org/wiki/Reciprocal_polynomial where the assertion from the current answer that '[the fact that $1/(-2) = 10 mod 7$] implies your claim immediately' is (in slightly more general notation) listed as 'Property 2.'
– Vincent
Aug 21 at 15:04
add a comment |Â
up vote
2
down vote
This is a trivial assertion if you understand modular arithmetic, which is absolutely basic in number theory.
Note that $1/(-2)equiv 10 pmod 7$, which implies your claim immediately.
However, this rule has disadvantages compared to the usual algorithm that decides if a number is divisible by $7$ or not (namely, break it up to $3$-digit numbers, add them with alternating signs, iterate...)
- Your method only decides if the number is divisible by $7$, it does not return the remainder $pmod 7$
- Your method is very slow, it takes much more time to compute that expression than adding up $3$-digit numbers.
So I suggest you read about advanced number theory, and if you find an interesting open problem (that experts found interesting as well), work on it.
But this result is unpublishable.
In any case, congratulations for finding this method.
answered Aug 21 at 13:20
A. Pongrácz
4,112625
I was hoping that the OP would take my hint & discover the 3 digit method for himself. But I guess it's still a good exercise to prove why it works.
– PM 2Ring
Aug 21 at 13:31
I think I have already finished my post (or maybe still typing it) when you made your suggestion, so I didn't see it. Sorry for that.
– A. Pongrácz
Aug 21 at 13:38
Not a problem! Your answer is fine.
– PM 2Ring
Aug 21 at 13:44
I am not sure if I would say it is trivial. The reversal of the order of the digits when passing from base 10 to base -2 is a nice twist which looks somewhat surprising. But it is true: things like these have been studied before in more generality. For instance you (that is: @Ravi Goyal, not A. Pongracz) could look at this page en.wikipedia.org/wiki/Reciprocal_polynomial where the assertion from the current answer that '[the fact that $1/(-2) = 10 mod 7$] implies your claim immediately' is (in slightly more general notation) listed as 'Property 2.'
– Vincent
Aug 21 at 15:04
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is a trivial assertion if you understand modular arithmetic, which is absolutely basic in number theory.
Note that $1/(-2)equiv 10 pmod 7$, which implies your claim immediately.
However, this rule has disadvantages compared to the usual algorithm that decides if a number is divisible by $7$ or not (namely, break it up to $3$-digit numbers, add them with alternating signs, iterate...)
- Your method only decides if the number is divisible by $7$, it does not return the remainder $pmod 7$
- Your method is very slow, it takes much more time to compute that expression than adding up $3$-digit numbers.
So I suggest you read about advanced number theory, and if you find an interesting open problem (that experts found interesting as well), work on it.
But this result is unpublishable.
In any case, congratulations for finding this method.
answered Aug 21 at 13:20
A. Pongrácz
4,112625
This is a trivial assertion if you understand modular arithmetic, which is absolutely basic in number theory.
Note that $1/(-2)equiv 10 pmod 7$, which implies your claim immediately.
However, this rule has disadvantages compared to the usual algorithm that decides if a number is divisible by $7$ or not (namely, break it up to $3$-digit numbers, add them with alternating signs, iterate...)
- Your method only decides if the number is divisible by $7$, it does not return the remainder $pmod 7$
- Your method is very slow, it takes much more time to compute that expression than adding up $3$-digit numbers.
So I suggest you read about advanced number theory, and if you find an interesting open problem (that experts found interesting as well), work on it.
But this result is unpublishable.
In any case, congratulations for finding this method.
answered Aug 21 at 13:20
A. Pongrácz
4,112625
answered Aug 21 at 13:20
A. Pongrácz
4,112625
answered Aug 21 at 13:20
A. Pongrácz
4,112625
answered Aug 21 at 13:20
A. Pongrácz
4,112625
4,112625
I was hoping that the OP would take my hint & discover the 3 digit method for himself. But I guess it's still a good exercise to prove why it works.
– PM 2Ring
Aug 21 at 13:31
I think I have already finished my post (or maybe still typing it) when you made your suggestion, so I didn't see it. Sorry for that.
– A. Pongrácz
Aug 21 at 13:38
Not a problem! Your answer is fine.
– PM 2Ring
Aug 21 at 13:44
I am not sure if I would say it is trivial. The reversal of the order of the digits when passing from base 10 to base -2 is a nice twist which looks somewhat surprising. But it is true: things like these have been studied before in more generality. For instance you (that is: @Ravi Goyal, not A. Pongracz) could look at this page en.wikipedia.org/wiki/Reciprocal_polynomial where the assertion from the current answer that '[the fact that $1/(-2) = 10 mod 7$] implies your claim immediately' is (in slightly more general notation) listed as 'Property 2.'
– Vincent
Aug 21 at 15:04
add a comment |Â
I was hoping that the OP would take my hint & discover the 3 digit method for himself. But I guess it's still a good exercise to prove why it works.
– PM 2Ring
Aug 21 at 13:31
I think I have already finished my post (or maybe still typing it) when you made your suggestion, so I didn't see it. Sorry for that.
– A. Pongrácz
Aug 21 at 13:38
Not a problem! Your answer is fine.
– PM 2Ring
Aug 21 at 13:44
I am not sure if I would say it is trivial. The reversal of the order of the digits when passing from base 10 to base -2 is a nice twist which looks somewhat surprising. But it is true: things like these have been studied before in more generality. For instance you (that is: @Ravi Goyal, not A. Pongracz) could look at this page en.wikipedia.org/wiki/Reciprocal_polynomial where the assertion from the current answer that '[the fact that $1/(-2) = 10 mod 7$] implies your claim immediately' is (in slightly more general notation) listed as 'Property 2.'
– Vincent
Aug 21 at 15:04
I was hoping that the OP would take my hint & discover the 3 digit method for himself. But I guess it's still a good exercise to prove why it works.
– PM 2Ring
Aug 21 at 13:31
I was hoping that the OP would take my hint & discover the 3 digit method for himself. But I guess it's still a good exercise to prove why it works.
– PM 2Ring
Aug 21 at 13:31
I think I have already finished my post (or maybe still typing it) when you made your suggestion, so I didn't see it. Sorry for that.
– A. Pongrácz
Aug 21 at 13:38
I think I have already finished my post (or maybe still typing it) when you made your suggestion, so I didn't see it. Sorry for that.
– A. Pongrácz
Aug 21 at 13:38
Not a problem! Your answer is fine.
– PM 2Ring
Aug 21 at 13:44
Not a problem! Your answer is fine.
– PM 2Ring
Aug 21 at 13:44
I am not sure if I would say it is trivial. The reversal of the order of the digits when passing from base 10 to base -2 is a nice twist which looks somewhat surprising. But it is true: things like these have been studied before in more generality. For instance you (that is: @Ravi Goyal, not A. Pongracz) could look at this page en.wikipedia.org/wiki/Reciprocal_polynomial where the assertion from the current answer that '[the fact that $1/(-2) = 10 mod 7$] implies your claim immediately' is (in slightly more general notation) listed as 'Property 2.'
– Vincent
Aug 21 at 15:04
I am not sure if I would say it is trivial. The reversal of the order of the digits when passing from base 10 to base -2 is a nice twist which looks somewhat surprising. But it is true: things like these have been studied before in more generality. For instance you (that is: @Ravi Goyal, not A. Pongracz) could look at this page en.wikipedia.org/wiki/Reciprocal_polynomial where the assertion from the current answer that '[the fact that $1/(-2) = 10 mod 7$] implies your claim immediately' is (in slightly more general notation) listed as 'Property 2.'
– Vincent
Aug 21 at 15:04
add a comment |Â
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1
What do $a,b,c,ldots$ represent? You haven't defined them. Without understanding your claim competely, I would suggest that you do not have your hopes up. Such rules are very well-understood for hundreds of years, these are not publishable.
– A. Pongrácz
Aug 21 at 13:11
If you believe the result is original, it could have been a good ideia to draft a paper and submit it to arxiv.org where some mathematicians publish time-sensitive results to ensure they're able to claim authorship. Then submit it to a good publication, where experts should check the validity and originality of your claim. Be careful that not much attention needs to be given if a poorly written paper is submitted. Good literature review is also important.
– Mefitico
Aug 21 at 13:12
Does this check only work for $7$-digit numbers?
– 5xum
Aug 21 at 13:15
It is as cumbersome as dividing a digit by $7$. Divisibility tests tend to make it easier like in the case of $2,3,11$ etc. Do you think this would be practical to solve for a $20$ digit number?
– prog_SAHIL
Aug 21 at 13:17
2
This is essentially the divisibility test on the wiki page where you double the last digit and subtract it from the number formed by the other digits. Then repeat the process with this new number, and so on until you have a single digit left. This is almost the same as multiplying the digits by powers of $-2$ as you are doing, but easier.
– Jaap Scherphuis
Aug 21 at 13:34